Coulomb's Law - SAT

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Question

If the distance between two charged particles is doubled, the strength of the electric force between them will __________.

Answer

Coulomb's law gives the relationship between the force of an electric field and the distance between two charges:

$F_e=k\frac{q_1q_2}{r^2}$

The strength of the force will be inversely proportional to the square of the distance between the charges.

When the distance between the charges is doubled, the total force will be divided by four (quartered).

$F_e=k\frac{q_1q_2}{(2r)^2}=k\frac{q_1q_2}{4r^2}=\frac{1}{4}(k\frac{q_1q_2}{r^2})$

Compare your answer with the correct one above

Question

If the distance between two charged particles is doubled, the strength of the electric force between them will __________.

Answer

Coulomb's law gives the relationship between the force of an electric field and the distance between two charges:

$F_e=k\frac{q_1q_2}{r^2}$

The strength of the force will be inversely proportional to the square of the distance between the charges.

When the distance between the charges is doubled, the total force will be divided by four (quartered).

$F_e=k\frac{q_1q_2}{(2r)^2}=k\frac{q_1q_2}{4r^2}=\frac{1}{4}(k\frac{q_1q_2}{r^2})$

Compare your answer with the correct one above

Question

If the distance between two charged particles is doubled, the strength of the electric force between them will __________.

Answer

Coulomb's law gives the relationship between the force of an electric field and the distance between two charges:

$F_e=k\frac{q_1q_2}{r^2}$

The strength of the force will be inversely proportional to the square of the distance between the charges.

When the distance between the charges is doubled, the total force will be divided by four (quartered).

$F_e=k\frac{q_1q_2}{(2r)^2}=k\frac{q_1q_2}{4r^2}=\frac{1}{4}(k\frac{q_1q_2}{r^2})$

Compare your answer with the correct one above

Question

If the distance between two charged particles is doubled, the strength of the electric force between them will __________.

Answer

Coulomb's law gives the relationship between the force of an electric field and the distance between two charges:

$F_e=k\frac{q_1q_2}{r^2}$

The strength of the force will be inversely proportional to the square of the distance between the charges.

When the distance between the charges is doubled, the total force will be divided by four (quartered).

$F_e=k\frac{q_1q_2}{(2r)^2}=k\frac{q_1q_2}{4r^2}=\frac{1}{4}(k\frac{q_1q_2}{r^2})$

Compare your answer with the correct one above

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